Search for a Range

Given an array of integers sorted in ascending order, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example, Given [5, 7, 7, 8, 8, 10] and target value 8, return [3, 4].

Solution: Using Binary Search

The key is to find the first index of target, and then find the first index of target+1.

    int searchRangeUtil(vector<int> A, int left, int right, int target) {
        while (left < right) {
            int mid = left + (right - left)/2;
            if (A[mid] < target) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return left;
    }

    vector<int> searchRange(vector<int> &A, int target) {
        int left = searchRangeUtil(A, 0, A.size(), target);
        int right = searchRangeUtil(A, left, A.size(), target+1);
        if (left > right-1) return {-1, -1};
        return {left, right-1};
    }